A force of 14.0N is applied to a block as shown in the diagram below


If the coefficient of friction is between the surface and the blocks is 0.35

a) What is the acceleration of the two blocks?

b) What is the force that the 3.0kg exerts on the 1.0kg block?

Solutions Given a)

$\sum F = 14.0N - 13.7N = 0.28N$

$\frac{0.28N}{4.0kg} = 0.07m/s^2 = a$


$\sum F = (1.0kg)(0.07m/s^2) = 0.070N$

$F_{fric} = (0.35)(1.0kg)(9.8m/s^2) = 3.43N$

$F = 0.07N + 3.43 = 3.5N$


Here is the thing, but my never show (probably due to print cost?) the free-body diagrams. I want to understand a systematic way of solving this, that is, draw FBD and then apply Newton's Second law. I already know that they first considered the whole system as "one mass" and they applied the 2nd Law and then the third law on each. I want to do this with a Free-Body Diagram.

Here is my attempt. I am going to denote the masses as 1 and 3 in the subscripts.

Here are my freebody diagrams.

For the block with mass 3kg


For the block with mass 1kg


My equations should be

$\sum F = F - (\mu m_3 g + F_{13}) = m_3 a$

$\sum F = F_{31} - \mu m_1 g = m_1 a$

Here is the problem, if I attempt to solve this I get (add the two equations together)

$F - \mu (m_1 + m_3)g - F_{13} + F_{31} = (m_1 + m_3)a $

If I were to go ahead and plug in the values I get

$14N - 13.72N - F_{13} + F_{31} = 4a $

$0.28N - F_{13} + F_{31} = 4a $

How come I can't get the same answer this way? Am I missing another equation? Perhaps a constraint? It would seem to make sense if it happens that $|F_{13}| - |F_{31}| = 0$

But when I broke it into components like this, I am dealing with the vectorial components, not the magnitiude. So really vectorically, I need $\vec{F}_{13} = -\vec{F}_{31}$

So I actually get if I plug in the vector force

Thank you

  • $\begingroup$ In the first equation after your FB diagrams you have positive $F_{31}$ pointing left. This setup does satisfy your condition you seem to doubt later. With your setup $F_{13} + F_{31} = 0$, and $F_{13} - F_{31} = 14$. $\endgroup$
    – Nic
    Dec 19, 2011 at 0:15
  • $\begingroup$ you should really use inkscape or gimp for drawing diagrams...forget about paint!! :P $\endgroup$ Apr 18, 2012 at 12:28
  • $\begingroup$ My first suggestion would be: Draw the coordinate system! Then, write the forces in vector form (this makes everything unambiguous). $\endgroup$ Dec 11, 2012 at 22:51
  • $\begingroup$ @glorfindel while I appreciate the effort to ensuring older questions will still be viewable now, each approved review pops these back into the main page & I'm not sure how useful it is to bring ~+1 score questions that are so old back to the front. $\endgroup$
    – Kyle Kanos
    Jan 12, 2019 at 0:44
  • $\begingroup$ @KyleKanos thanks for reaching out to me. I've taken some steps to ensure it doesn't happen too often (1 suggested edit per 36 hours, sometimes two additional edits to Community Wiki posts). As a comparison, the Community user bumps questions much more often. On a volume of 80 active questions per day on Physics, I'm surprised somebody noticed :) This project of mine is generally well-received across the Stack Exchange network; for some more data and discussion, see meta.mathoverflow.net/a/4058/70594 $\endgroup$
    – Glorfindel
    Jan 12, 2019 at 7:58

2 Answers 2


$$\vec{F}_{13} = -\vec{F}_{31}$$

This is nothing more or less than the precise mathematical statement of Newton's third law for objects 1 and 3. In words, it reads

the force exerted by object 1 on object 3 is equal (in magnitude) and opposite (in direction) to the force exerted by object 3 on object 1

So I think you've got it ;-)

Keep in mind that when you are drawing free body diagrams for two objects that exert forces on each other, you will always get a matched pair of arrows, one in each diagram, because of Newton's third law. In this case, the arrow corresponding to $\vec{F}_{13}$ (in the diagram for $m_3$) and the one corresponding to $\vec{F}_{31}$ (in the diagram for $m_1$) are your matched pair. This may be obvious to you, but I just mention it because when I teach this topic to intro physics students, a lot of them have a hard time realizing that Newton's third law requires those matched pairs of arrows.

  • $\begingroup$ Do people normally solve problems like this with my method? I am guessing not. It would considered "awkward" right? $\endgroup$
    – Lemon
    Dec 19, 2011 at 2:13
  • $\begingroup$ Yes, people do. Awkward or not, your method is the correct way to do it and is exactly what I teach in my intro physics sections. $\endgroup$
    – David Z
    Dec 19, 2011 at 2:45

Look at D'Alenberg's principle which replaces each known/unknown acceleration, with and equal and opposite force of $\vec{F} = -m\,\vec{a}$ such that the sum of the forces equal zero for each rigid body. Do one FBD for each body and force the two blocks to move together. There are two equations, one for each FBD, and two unknowns the motion and the contact force.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.